The solution of the equation cos^2 x - 2cos x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\cos^2 x - 2\cos x = 4\sin x - 2\sin x \cos x$$\c

Vedclass · Accepted Answer

$\pi + 	an^{-1}\left(-\frac{1}{2}ight)$

Vedclass · Answer

(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\cos^2 x - 2\cos x = 4\sin x - 2\sin x \cos x$$\cos x(\cos x - 2) = 2\sin x(2 - \cos x)$$\cos x(\cos x - 2) + 2\sin x(\cos x - 2) = 0$$(\cos x - 2)(\cos x + 2\sin x) = 0$Since $\cos x - 2 
eq 0$ for any real $x$,we must have $\cos x + 2\sin x = 0$.$	an x = -\frac{1}{2}$Since $0 \le x \le \pi$,and $	an x$ is negative in the second quadrant,the solution is $x = \pi + 	an^{-1}\left(-\frac{1}{2}ight)$.

The solution of the equation $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$ for $0 \le x \le \pi$ is

Similar Questions

The principal solutions of $\cot x = \sqrt{3}$ are

Find the general solution of the equation $\sin x + \sin 3x + \sin 5x = 0$.

The number of principal solutions of the equation $\tan 3x - \tan 2x - \tan x = 0$ is:

The number of solutions of $\sin x + \sin 3x + \sin 5x = 0$ in the interval $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ is

What is the number of roots of the quadratic equation $8\sec^2\theta - 6\sec\theta + 1 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module